Question 1: Consider the following reaction: SO2Cl2(g)SO2(g)+Cl2(g) K c=2.99×107
ID: 1081732 • Letter: Q
Question
Question 1: Consider the following reaction:
SO2Cl2(g)SO2(g)+Cl2(g)
Kc=2.99×107 at 227 C
If a reaction mixture initially contains 0.195 MSO2Cl2, what is the equilibrium concentration of Cl2at 227 C??
Question 2:
Consider the reaction
CO(g)+H2O(g)CO2(g)+H2(g)
Kc=102 at 500 K
A reaction mixture initially contains 0.130 MCOand 0.130 MH2O.
What will be the equilibrium concentration of H2O?
Question 3:
The following reaction was performed in a sealed vessel at 768 C :
H2(g)+I2(g)2HI(g)
Initially, only H2 and I2 were present at concentrations of [H2]=3.65M and [I2]=2.70M. The equilibrium concentration of I2 is 0.0800 M . What is the equilibrium constant, Kc, for the reaction at this temperature?
Please explain and answer all! Thank you
Explanation / Answer
Question 1: Consider the following reaction:
SO2Cl2(g)SO2(g)+Cl2(g)
Kc=2.99×107 at 227 C
If a reaction mixture initially contains 0.195 MSO2Cl2, what is the equilibrium concentration of Cl2at 227 C??
Answer :-
Given reaction is in equilibrium,
so we can write
SOCl2
SO2
Cl2
Initial Concentration
0.195
0
0
Change in concentration
-x
+x
+x
Concentration After Equilibrium
0.195 -x
+x
+x
We can write an equation of reaction concentration as
K = [SO2][O2]/[SO2Cl2]
Lets fill the values from table in the formula
2.99 x 10^-7 = (x)(x)/(0.195 - x )
Neglect x in the 0.195 -x ( because X is very small, so lets consider it as only 0.195)
2.99 x 10^-7 = (x)(x)/0.195
x^2 = 5.8305*10^-8
x = 2.4146 *10^-4
X is nothing but the concentration of Cl2 gas
So, x = 2.18 x 10-4 M is our final answer.
Consider the reaction
CO(g)+H2O(g)CO2(g)+H2(g)
Kc=102 at 500 K
A reaction mixture initially contains 0.130 MCOand 0.130 MH2O.
What will be the equilibrium concentration of H2O?
Answer :-
Given reaction is in equilibrium,
so we can write
CO(g)
H2O(g)
CO2(g)
H2(g)
Initial Concentration
0.130 M
0.130 M
0
0
Change in concentration
-x
-x
+x
+x
Concentration After Equilibrium
0.130 -x
0.130 -x
+x
+x
We can write an equation of reaction concentration as
Ka = [CO2] [H2] / ([CO] [H2O])
Lets put the values in the formula
102 = (X)(X)/ (0.130 –x)*( 0.130 –x)
102 = (X)^2 / ( 0.130 –x)^2
Lest take a square root of this equation
10.10 = (X) / ( 0.130 –x)
10.10*(0.130 –x) = (X)
1.3130- 10.10 X= X
1.3130 = X + 10.10 X
1.3130 = 11.10 X
X = (1.3130)/ 11.10
X= 0.1183 M
Equilibrium concentration of H2O = 0.130- X
= 0.130-0.1183 M
= 0.0117 M
This is our final answer
The following reaction was performed in a sealed vessel at 768 C :
H2(g)+I2(g)2HI(g)
Initially, only H2 and I2 were present at concentrations of [H2]=3.65M and [I2]=2.70M. The equilibrium concentration of I2 is 0.0800 M . What is the equilibrium constant, Kc, for the reaction at this temperature?
Answer :-
Given reaction is in equilibrium,
so we can write
H2(g)+I2(g)2HI(g)
As per the given reaction
When 1 mole of I2 reacts with 1 mole of H2 then 2 moles of product (HI) is formed
So when 0.08 M of I2 reacts with 0.08 M of H2 then 1.6M (2*0.08M) of HI forms
H2
I2
2HI
Initial Concentration
3.65
2.70
0
Change in concentration
0.0800
0.0800(given )
1.6
Concentration After Equilibrium
(3.65-0.0800) =3.57
(2.70-0.0800)=
2.62
1.6
We can write an equation of reaction concentration as
K = [HI]^2/[H2] [I2]
Lets fill the values from table in the formula
K = [1.6]^2/[3.57] [2.62]
lets solve this for K
K = [1.6]^2/[3.57] [2.62]
K = [2.56]/[9.3534]
K =0.27370
SOCl2
SO2
Cl2
Initial Concentration
0.195
0
0
Change in concentration
-x
+x
+x
Concentration After Equilibrium
0.195 -x
+x
+x
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